\(\int \frac {8 x-6 x^2-2 x^3+(-8 x-2 x^2) \log (4+x)+e^{e^x-x^2 \log (-1+x+\log (4+x))} (-4+3 x+x^2-5 x^3-x^4+e^x (-4 x+3 x^2+x^3)+(4+x+e^x (4 x+x^2)) \log (4+x)+(8 x^2-6 x^3-2 x^4+(-8 x^2-2 x^3) \log (4+x)) \log (-1+x+\log (4+x)))}{4 x^2-3 x^3-x^4+(-4 x^2-x^3) \log (4+x)+e^{e^x-x^2 \log (-1+x+\log (4+x))} (-4 x+3 x^2+x^3+(4 x+x^2) \log (4+x))} \, dx\) [8812]

Optimal result

Integrand size = 221, antiderivative size = 26 \[ \int \frac {8 x-6 x^2-2 x^3+\left (-8 x-2 x^2\right ) \log (4+x)+e^{e^x-x^2 \log (-1+x+\log (4+x))} \left (-4+3 x+x^2-5 x^3-x^4+e^x \left (-4 x+3 x^2+x^3\right )+\left (4+x+e^x \left (4 x+x^2\right )\right ) \log (4+x)+\left (8 x^2-6 x^3-2 x^4+\left (-8 x^2-2 x^3\right ) \log (4+x)\right ) \log (-1+x+\log (4+x))\right )}{4 x^2-3 x^3-x^4+\left (-4 x^2-x^3\right ) \log (4+x)+e^{e^x-x^2 \log (-1+x+\log (4+x))} \left (-4 x+3 x^2+x^3+\left (4 x+x^2\right ) \log (4+x)\right )} \, dx=\log \left (x \left (-e^{e^x-x^2 \log (-1+x+\log (4+x))}+x\right )\right ) \]

[Out]

ln(x*(x-exp(-x^2*ln(ln(4+x)+x-1)+exp(x))))

Rubi [F(-1)]

Timed out. \[ \int \frac {8 x-6 x^2-2 x^3+\left (-8 x-2 x^2\right ) \log (4+x)+e^{e^x-x^2 \log (-1+x+\log (4+x))} \left (-4+3 x+x^2-5 x^3-x^4+e^x \left (-4 x+3 x^2+x^3\right )+\left (4+x+e^x \left (4 x+x^2\right )\right ) \log (4+x)+\left (8 x^2-6 x^3-2 x^4+\left (-8 x^2-2 x^3\right ) \log (4+x)\right ) \log (-1+x+\log (4+x))\right )}{4 x^2-3 x^3-x^4+\left (-4 x^2-x^3\right ) \log (4+x)+e^{e^x-x^2 \log (-1+x+\log (4+x))} \left (-4 x+3 x^2+x^3+\left (4 x+x^2\right ) \log (4+x)\right )} \, dx=\text {\$Aborted} \]

[In]

Int[(8*x - 6*x^2 - 2*x^3 + (-8*x - 2*x^2)*Log[4 + x] + E^(E^x - x^2*Log[-1 + x + Log[4 + x]])*(-4 + 3*x + x^2
- 5*x^3 - x^4 + E^x*(-4*x + 3*x^2 + x^3) + (4 + x + E^x*(4*x + x^2))*Log[4 + x] + (8*x^2 - 6*x^3 - 2*x^4 + (-8
*x^2 - 2*x^3)*Log[4 + x])*Log[-1 + x + Log[4 + x]]))/(4*x^2 - 3*x^3 - x^4 + (-4*x^2 - x^3)*Log[4 + x] + E^(E^x
- x^2*Log[-1 + x + Log[4 + x]])*(-4*x + 3*x^2 + x^3 + (4*x + x^2)*Log[4 + x])),x]

[Out]

$Aborted

Rubi steps Aborted

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {8 x-6 x^2-2 x^3+\left (-8 x-2 x^2\right ) \log (4+x)+e^{e^x-x^2 \log (-1+x+\log (4+x))} \left (-4+3 x+x^2-5 x^3-x^4+e^x \left (-4 x+3 x^2+x^3\right )+\left (4+x+e^x \left (4 x+x^2\right )\right ) \log (4+x)+\left (8 x^2-6 x^3-2 x^4+\left (-8 x^2-2 x^3\right ) \log (4+x)\right ) \log (-1+x+\log (4+x))\right )}{4 x^2-3 x^3-x^4+\left (-4 x^2-x^3\right ) \log (4+x)+e^{e^x-x^2 \log (-1+x+\log (4+x))} \left (-4 x+3 x^2+x^3+\left (4 x+x^2\right ) \log (4+x)\right )} \, dx=\log (x)-x^2 \log (-1+x+\log (4+x))+\log \left (e^{e^x}-x (-1+x+\log (4+x))^{x^2}\right ) \]

[In]

Integrate[(8*x - 6*x^2 - 2*x^3 + (-8*x - 2*x^2)*Log[4 + x] + E^(E^x - x^2*Log[-1 + x + Log[4 + x]])*(-4 + 3*x
+ x^2 - 5*x^3 - x^4 + E^x*(-4*x + 3*x^2 + x^3) + (4 + x + E^x*(4*x + x^2))*Log[4 + x] + (8*x^2 - 6*x^3 - 2*x^4
+ (-8*x^2 - 2*x^3)*Log[4 + x])*Log[-1 + x + Log[4 + x]]))/(4*x^2 - 3*x^3 - x^4 + (-4*x^2 - x^3)*Log[4 + x] +
E^(E^x - x^2*Log[-1 + x + Log[4 + x]])*(-4*x + 3*x^2 + x^3 + (4*x + x^2)*Log[4 + x])),x]

[Out]

Log[x] - x^2*Log[-1 + x + Log[4 + x]] + Log[E^E^x - x*(-1 + x + Log[4 + x])^x^2]

Maple [A] (verified)

Time = 242.35 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00

method	result	size
risch	\(\ln \left (x \right )+\ln \left (\left (\ln \left (4+x \right )+x -1\right )^{-x^{2}} {\mathrm e}^{{\mathrm e}^{x}}-x \right )\)	\(26\)
parallelrisch	\(\ln \left (x \right )+\ln \left (x -{\mathrm e}^{-x^{2} \ln \left (\ln \left (4+x \right )+x -1\right )+{\mathrm e}^{x}}\right )\)	\(26\)

[In]

int(((((-2*x^3-8*x^2)*ln(4+x)-2*x^4-6*x^3+8*x^2)*ln(ln(4+x)+x-1)+((x^2+4*x)*exp(x)+4+x)*ln(4+x)+(x^3+3*x^2-4*x
)*exp(x)-x^4-5*x^3+x^2+3*x-4)*exp(-x^2*ln(ln(4+x)+x-1)+exp(x))+(-2*x^2-8*x)*ln(4+x)-2*x^3-6*x^2+8*x)/(((x^2+4*
x)*ln(4+x)+x^3+3*x^2-4*x)*exp(-x^2*ln(ln(4+x)+x-1)+exp(x))+(-x^3-4*x^2)*ln(4+x)-x^4-3*x^3+4*x^2),x,method=_RET
URNVERBOSE)

[Out]

ln(x)+ln((ln(4+x)+x-1)^(-x^2)*exp(exp(x))-x)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {8 x-6 x^2-2 x^3+\left (-8 x-2 x^2\right ) \log (4+x)+e^{e^x-x^2 \log (-1+x+\log (4+x))} \left (-4+3 x+x^2-5 x^3-x^4+e^x \left (-4 x+3 x^2+x^3\right )+\left (4+x+e^x \left (4 x+x^2\right )\right ) \log (4+x)+\left (8 x^2-6 x^3-2 x^4+\left (-8 x^2-2 x^3\right ) \log (4+x)\right ) \log (-1+x+\log (4+x))\right )}{4 x^2-3 x^3-x^4+\left (-4 x^2-x^3\right ) \log (4+x)+e^{e^x-x^2 \log (-1+x+\log (4+x))} \left (-4 x+3 x^2+x^3+\left (4 x+x^2\right ) \log (4+x)\right )} \, dx=\log \left (x\right ) + \log \left (-x + e^{\left (-x^{2} \log \left (x + \log \left (x + 4\right ) - 1\right ) + e^{x}\right )}\right ) \]

[In]

integrate(((((-2*x^3-8*x^2)*log(4+x)-2*x^4-6*x^3+8*x^2)*log(log(4+x)+x-1)+((x^2+4*x)*exp(x)+4+x)*log(4+x)+(x^3
+3*x^2-4*x)*exp(x)-x^4-5*x^3+x^2+3*x-4)*exp(-x^2*log(log(4+x)+x-1)+exp(x))+(-2*x^2-8*x)*log(4+x)-2*x^3-6*x^2+8
*x)/(((x^2+4*x)*log(4+x)+x^3+3*x^2-4*x)*exp(-x^2*log(log(4+x)+x-1)+exp(x))+(-x^3-4*x^2)*log(4+x)-x^4-3*x^3+4*x
^2),x, algorithm="fricas")

[Out]

log(x) + log(-x + e^(-x^2*log(x + log(x + 4) - 1) + e^x))

Sympy [A] (verification not implemented)

Time = 2.74 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {8 x-6 x^2-2 x^3+\left (-8 x-2 x^2\right ) \log (4+x)+e^{e^x-x^2 \log (-1+x+\log (4+x))} \left (-4+3 x+x^2-5 x^3-x^4+e^x \left (-4 x+3 x^2+x^3\right )+\left (4+x+e^x \left (4 x+x^2\right )\right ) \log (4+x)+\left (8 x^2-6 x^3-2 x^4+\left (-8 x^2-2 x^3\right ) \log (4+x)\right ) \log (-1+x+\log (4+x))\right )}{4 x^2-3 x^3-x^4+\left (-4 x^2-x^3\right ) \log (4+x)+e^{e^x-x^2 \log (-1+x+\log (4+x))} \left (-4 x+3 x^2+x^3+\left (4 x+x^2\right ) \log (4+x)\right )} \, dx=\log {\left (x \right )} + \log {\left (- x + e^{- x^{2} \log {\left (x + \log {\left (x + 4 \right )} - 1 \right )} + e^{x}} \right )} \]

[In]

integrate(((((-2*x**3-8*x**2)*ln(4+x)-2*x**4-6*x**3+8*x**2)*ln(ln(4+x)+x-1)+((x**2+4*x)*exp(x)+4+x)*ln(4+x)+(x
**3+3*x**2-4*x)*exp(x)-x**4-5*x**3+x**2+3*x-4)*exp(-x**2*ln(ln(4+x)+x-1)+exp(x))+(-2*x**2-8*x)*ln(4+x)-2*x**3-
6*x**2+8*x)/(((x**2+4*x)*ln(4+x)+x**3+3*x**2-4*x)*exp(-x**2*ln(ln(4+x)+x-1)+exp(x))+(-x**3-4*x**2)*ln(4+x)-x**
4-3*x**3+4*x**2),x)

[Out]

log(x) + log(-x + exp(-x**2*log(x + log(x + 4) - 1) + exp(x)))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {8 x-6 x^2-2 x^3+\left (-8 x-2 x^2\right ) \log (4+x)+e^{e^x-x^2 \log (-1+x+\log (4+x))} \left (-4+3 x+x^2-5 x^3-x^4+e^x \left (-4 x+3 x^2+x^3\right )+\left (4+x+e^x \left (4 x+x^2\right )\right ) \log (4+x)+\left (8 x^2-6 x^3-2 x^4+\left (-8 x^2-2 x^3\right ) \log (4+x)\right ) \log (-1+x+\log (4+x))\right )}{4 x^2-3 x^3-x^4+\left (-4 x^2-x^3\right ) \log (4+x)+e^{e^x-x^2 \log (-1+x+\log (4+x))} \left (-4 x+3 x^2+x^3+\left (4 x+x^2\right ) \log (4+x)\right )} \, dx=-x^{2} \log \left (x + \log \left (x + 4\right ) - 1\right ) + 2 \, \log \left (x\right ) + \log \left (\frac {{\left (x + \log \left (x + 4\right ) - 1\right )}^{\left (x^{2}\right )} x - e^{\left (e^{x}\right )}}{x}\right ) \]

[In]

integrate(((((-2*x^3-8*x^2)*log(4+x)-2*x^4-6*x^3+8*x^2)*log(log(4+x)+x-1)+((x^2+4*x)*exp(x)+4+x)*log(4+x)+(x^3
+3*x^2-4*x)*exp(x)-x^4-5*x^3+x^2+3*x-4)*exp(-x^2*log(log(4+x)+x-1)+exp(x))+(-2*x^2-8*x)*log(4+x)-2*x^3-6*x^2+8
*x)/(((x^2+4*x)*log(4+x)+x^3+3*x^2-4*x)*exp(-x^2*log(log(4+x)+x-1)+exp(x))+(-x^3-4*x^2)*log(4+x)-x^4-3*x^3+4*x
^2),x, algorithm="maxima")

[Out]

-x^2*log(x + log(x + 4) - 1) + 2*log(x) + log(((x + log(x + 4) - 1)^(x^2)*x - e^(e^x))/x)

Giac [A] (verification not implemented)

none

Time = 9.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {8 x-6 x^2-2 x^3+\left (-8 x-2 x^2\right ) \log (4+x)+e^{e^x-x^2 \log (-1+x+\log (4+x))} \left (-4+3 x+x^2-5 x^3-x^4+e^x \left (-4 x+3 x^2+x^3\right )+\left (4+x+e^x \left (4 x+x^2\right )\right ) \log (4+x)+\left (8 x^2-6 x^3-2 x^4+\left (-8 x^2-2 x^3\right ) \log (4+x)\right ) \log (-1+x+\log (4+x))\right )}{4 x^2-3 x^3-x^4+\left (-4 x^2-x^3\right ) \log (4+x)+e^{e^x-x^2 \log (-1+x+\log (4+x))} \left (-4 x+3 x^2+x^3+\left (4 x+x^2\right ) \log (4+x)\right )} \, dx=\log \left (x - e^{\left (-x^{2} \log \left (x + \log \left (x + 4\right ) - 1\right ) + e^{x}\right )}\right ) + \log \left (x\right ) \]

[In]

integrate(((((-2*x^3-8*x^2)*log(4+x)-2*x^4-6*x^3+8*x^2)*log(log(4+x)+x-1)+((x^2+4*x)*exp(x)+4+x)*log(4+x)+(x^3
+3*x^2-4*x)*exp(x)-x^4-5*x^3+x^2+3*x-4)*exp(-x^2*log(log(4+x)+x-1)+exp(x))+(-2*x^2-8*x)*log(4+x)-2*x^3-6*x^2+8
*x)/(((x^2+4*x)*log(4+x)+x^3+3*x^2-4*x)*exp(-x^2*log(log(4+x)+x-1)+exp(x))+(-x^3-4*x^2)*log(4+x)-x^4-3*x^3+4*x
^2),x, algorithm="giac")

[Out]

log(x - e^(-x^2*log(x + log(x + 4) - 1) + e^x)) + log(x)

Mupad [B] (verification not implemented)

Time = 14.81 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {8 x-6 x^2-2 x^3+\left (-8 x-2 x^2\right ) \log (4+x)+e^{e^x-x^2 \log (-1+x+\log (4+x))} \left (-4+3 x+x^2-5 x^3-x^4+e^x \left (-4 x+3 x^2+x^3\right )+\left (4+x+e^x \left (4 x+x^2\right )\right ) \log (4+x)+\left (8 x^2-6 x^3-2 x^4+\left (-8 x^2-2 x^3\right ) \log (4+x)\right ) \log (-1+x+\log (4+x))\right )}{4 x^2-3 x^3-x^4+\left (-4 x^2-x^3\right ) \log (4+x)+e^{e^x-x^2 \log (-1+x+\log (4+x))} \left (-4 x+3 x^2+x^3+\left (4 x+x^2\right ) \log (4+x)\right )} \, dx=\ln \left (x\right )+\ln \left (\frac {{\mathrm {e}}^{{\mathrm {e}}^x}}{{\left (x+\ln \left (x+4\right )-1\right )}^{x^2}}-x\right ) \]

[In]

int((log(x + 4)*(8*x + 2*x^2) - 8*x + 6*x^2 + 2*x^3 - exp(exp(x) - x^2*log(x + log(x + 4) - 1))*(3*x - log(x +
log(x + 4) - 1)*(log(x + 4)*(8*x^2 + 2*x^3) - 8*x^2 + 6*x^3 + 2*x^4) + exp(x)*(3*x^2 - 4*x + x^3) + log(x + 4
)*(x + exp(x)*(4*x + x^2) + 4) + x^2 - 5*x^3 - x^4 - 4))/(log(x + 4)*(4*x^2 + x^3) - exp(exp(x) - x^2*log(x +
log(x + 4) - 1))*(log(x + 4)*(4*x + x^2) - 4*x + 3*x^2 + x^3) - 4*x^2 + 3*x^3 + x^4),x)

[Out]

log(x) + log(exp(exp(x))/(x + log(x + 4) - 1)^(x^2) - x)